 # Composition Of Functions & Inverse Of A Function

## Composite Functions

When two functions are combined in such a way that the output of one function becomes the input to another function, then this is referred to as composite function. Consider three sets X, Y and Z and let f : X → Y  and g: Y → Z.

According to this, under map f, an element  x ∈ X is mapped to an element y = f(x) ∈ Y which in turn is mapped by g to an element z ∈ Z  in such a manner that z = g(y) = g[(f(x)] .

This mapping comprising of mappings f and g is known as composition of mappings. It is denoted by gof . Therefore, we are mapping onto .

The composite function is denoted as:

$~~~~~~~~~$ ( gof)(x) = g(f (X) )

Similarly, (fog) (x) = f (g(x))

So, to find (gof) (x), take f(x)  as argument for the function g.

Let us try to solve some questions based on composite functions.

 Let’s Work Out: Example: Given the function f(x) = 3x + 5  and g(x) = $2x^3$  .Find ( gof)(x) and ( fog)(x). Solution:  We know, (gof) (x) = g(f(x)) = g (3x+5) = $2(3x+5)^3$ Using Binomial Expansion, we have $(gof) (x) = 2 \left [ (3x)^{3} + 3.(3x)^{2}.5 + 3.(3x)(5)^{2}+(5)^3 \right ]$ $\Rightarrow (gof) (x) = 2 \left [ 27x^{3} + 135x^{2} + 225 x + 125 \right]$ $\Rightarrow (gof) (x) = 54x^{3} + 270x^{2} + 450 x + 250$ Now, (fog)(x) = f $\left( g(x) \right)$ = f ( $2x^3$) = $3 (2x^{3}) + 5$ $\Rightarrow (fog)(x) = 6x^{3} + 5$ Example: Let f(x) = $x^2$ and g(x) = $\sqrt{1 – x^2}$ .Find (gof)(x) and ( fog)(x) . Solution: (gof)(x) = g( f(x)) = g($x^2$) = $\sqrt{1 – (x^2)^2}$ = $\sqrt{1 – x^4}$ (fog) (x) = f(g(x)) = f( $\sqrt{1 – x^2}$) = ( $\sqrt{1 – x^2})^2$ = $1 – x^2$

## Inverse of a Function

Let f:X → Y. Now, let f represent a one to one function and y be any element of Y , there exists a unique element x ∈ X  such that y = f(x).Then the map

$~~~~~~~~~~$ $f^{-1}:f[X] \rightarrow X$

which associates to each element is called as the inverse map of f.

The function f(x) = $x^5$ and g(x) = $x^{\frac 12}$   have the following property:

$f\left( g(x) \right)$ = $f \left( x^{\frac 15} \right)$ = $(x^{\frac 15} )^5$ = x

$g\left( f(x) \right)$ = $g \left( x^{5} \right)$ = $(x^{5} )^{\frac 15}$ = x

Thus, if two functions f and g  satisfy $f \left( g(x) \right)$ = x for every x in domain of f , then in such a situation we can say that the function f is the inverse of g  and g  is the inverse of f .

For finding the inverse of a function,we write down the function y as a function of x  i.e. y = f(x)   and then solve for x  as a function of y.

To have a better insight on the topic let us go through some examples.

 Let’s Work Out: Example: If f(x) = $x^2$, g(x) = $\frac{x}{3}$   and h(x) = 3x+2 . Find out fohog(x). Solution:  $h\left( g(x) \right)$ = 3 $\left ( \frac x3 \right)$ + 2 = x + 2 fohog(x) = f $\left[ h\left(g(x)\right)\right]$ = $( x+ 2)^2$ This is the required solution. Example: Example 2: Find the inverse of the function f(x) = $x^3$ , x ∈ R. Solution: The given function f(x) = $x^3$   is a one to one and onto function defined in the range → R  . Therefore, we can find the inverse of this function. To find the inverse, we need to write down this function as $~~~~~~~~~~~~~$ y = $x^3$ In the above equation,y  is an arbitrary element from the range of f. If we solve for x from the above equation, we will get: $~~~~~~~~~~~~~$ x = $y^{\frac 13}$ This gives a function g:Y →X. This new function g can be defined as $~~~~~~~~~~~~~$ g(y) = $y^{\frac 13}$ This function g is the inverse of the function f since it’s domain is same as the range of the function f.Since g(y) = $y ^{\frac 12}$ , representing the independent variable with x , we get g(x) = $x^{\frac 13}$ = $f^{-1}(x)$..